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Anit [1.1K]
3 years ago
12

An atom of radium (Ra) forms a monatomic ion.

Chemistry
1 answer:
Agata [3.3K]3 years ago
3 0

Answer:

The atomic number of the element Radium (Ra) is 88. It is an alkaline earth metal. Electronic configuration of Radium (Ra) will be [Rn]7s^{2}[Rn]7s

2

It belongs to the group IIA of the periodic table. Alkaline earth metals that belong to the group IIA of the periodic table lose the valence electrons to form a +2 ion.

Therefore, Radium with an electronic configuration of [Rn]7s^{2}[Rn]7s

2

will lose the two 7s electrons to form the cation with +2 charge.

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What are bonds in metals like?<br> (K12)
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Metallic bonds are the force of attraction between positive metal ions and the valence electrons that are constantly moving around them. The ions form a lattice-like structure held together by the metallic bonds. Metallic bonds explain why metals can conduct electricity and bend without breaking.

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Blast furnaces extract pure iron from the iron(III) oxide in iron ore in a two step sequence.In the first step, carbon and oxyge
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Answer:

The net chemical equation for the production of iron from carbon, oxygen and iron(III) oxide:

2Fe_2O_3 (s) +6C (s) + 3O_2 (g)\rightarrow 4Fe (s) + 6CO_2 (g)

Explanation:

Step 1: Carbon and oxygen react to form carbon monoxide:

2C (s) + O_2 (g)\rightarrow 2CO (g)..[1]

Step 2: Iron(III) oxide and carbon monoxide react to form iron and carbon dioxide:

Fe_2O_3 (s) + 3CO (g)\rightarrow 2Fe (s) + 3CO_2 (g)..[2]

The net chemical equation for the production of iron from carbon, oxygen and iron(III) oxide can be obtained by :

3 × [1] + 2 × [2]

2Fe_2O_3 (s) + 6CO (g)+6C (s) + 3O_2 (g)\rightarrow 4Fe (s) + 6CO_2 (g)+6CO(g)

CO gas will gas cancel out form both sides ans we will get the net chemical equation:

2Fe_2O_3 (s) +6C (s) + 3O_2 (g)\rightarrow 4Fe (s) + 6CO_2 (g)

According to reaction, 2 moles of ferric oxide reacts with 6 moles of carbon and 3 moles of oxygen gas to give 4 moles of iron and 6 moles pf carbon dioxide gas.

8 0
4 years ago
A laboratory analysis of a 100 g sample finds it is composed of 1.8 g hydrogen, 56.1 g sulfur, and 42.1 g oxygen. What is its em
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Answer: The empirical formula is H_2S_2O_3

Explanation:

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mas of H = 1.8 g

Mass of S = 56.1 g

Mass of O = 42.1 g

Step 1 : convert given masses into moles.

Moles of H =\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{1.8g}{1g/mole}=1.8moles

Mass of S =\frac{\text{ given mass of S}}{\text{ molar mass of S}}= \frac{56.1g}{32g/mole}=1.8moles

Moles of O=\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{42.1g}{16g/mole}=2.6moles

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For H = \frac{1.8}{1.8}=1

For S = \frac{1.8}{1.8}=1

For O =\frac{2.6}{1.8}=1.5

Converting to whole number ratios

The ratio of H: S: O= 2: 2: 3

Hence the empirical formula is H_2S_2O_3

7 0
3 years ago
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